杭电1004-Let the Balloon Rise

最新推荐文章于 2021-06-24 22:53:24 发布

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本文深入探讨了数据处理和算法优化的关键技术，包括数据结构选择、算法设计、性能评估等，旨在帮助开发者提升代码效率，解决实际问题。

Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 91657 Accepted Submission(s): 34895

Problem Description

Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color and find the result.

This year, they decide to leave this lovely job to you.

Input

Input contains multiple test cases. Each test case starts with a number N (0 < N <= 1000) -- the total number of balloons distributed. The next N lines contain one color each. The color of a balloon is a string of up to 15 lower-case letters.

A test case with N = 0 terminates the input and this test case is not to be processed.

Output

For each case, print the color of balloon for the most popular problem on a single line. It is guaranteed that there is a unique solution for each test case.

Sample Input

5
green
red
blue
red
red
3
pink
orange
pink
0

Sample Output

red
pink

直接暴力比较，找出重复次数最大的！

AC代码：

#include<cstdio>
#include<cstring>
int main()
{
	int N,i,j,cot,a[1100],best;
	char s[1010][20];
	while(scanf("%d",&N),N)
	{
		memset(a,0,sizeof(a));
		best=0;
		for(i=0;i<N;i++)
			scanf("%s",s[i]);
		for(i=0;i<N;i++)
		{
			for(j=i;j<N;j++)
			{
				if(strcmp(s[i],s[j])==0)
				a[i]++; 
			 } 
			 if(a[i]>best)
			  	 best=a[i]; 
		}
		for(i=0;i<N;i++)
		{
			if(a[i]==best)
			{
				printf("%s\n",s[i]);
				break;
			 } 
		 } 	 
	}
	return 0;
 }